What is the slope of the line tangent to $f(x) = 2x^{2}+x-5$ at $x = 1$ ?
Answer: The slope of the tangent line is $ \lim_{h \to 0} \frac{f(x + h) - f(x)}{h}$ $ = \lim_{h \to 0} \frac{(2(x+h)^{2}+x+h-5) - (2x^{2}+x-5)}{h}$ $ = \lim_{h \to 0} \frac{(2(x^{2}+2x h+h^{2})+x+h-5) - (2x^{2}+x-5)}{h}$ $ = \lim_{h \to 0} \frac{2x^{2}+4(x h)+2h^{2}+x+h-5-2x^{2}-x+5}{h}$ $ = \lim_{h \to 0} \frac{4(x h)+2h^{2}+h}{h}$ $ = \lim_{h \to 0} 4x+2h+1$ $ = 4x+1$ $ = (4)(1)+1$ $ = 5$